The graph shows the force exerted on an object versus the position of that object along the x-axis. The force has no components other than along the x-axis. If z is an increment of 25, then what is the work done by the force on the object as the object is displaced from 0 to 3 m?

To find the work done by the force on the object as it is displaced from 0 to 3 meters, we need to calculate the area under the graph of force versus position.

Since the force has no components other than along the x-axis, the work done can be calculated as the area under the graph of force versus position.

Let's divide the displacement into smaller increments of size z = 25. We will calculate the work done for each increment and then sum them up.

The total work done (W) is given by the formula:

W = ∑(F * Δx)

where ∑ represents the summation, F is the force at each position, and Δx is the displacement increment.

In this case, the displacement increments are:

Δx = 0 - 25 - 50 - 75 - 100 - 125 - 150 - 175 - 200 - 225 - 250 - 275 - 300

We will calculate the work done at each increment and then sum them up.

At position 0, the force is 0, so the work done is:

W0 = 0 * 25 = 0

At position 25, the force is 4 N, so the work done is:

W25 = 4 * 25 = 100

At position 50, the force is 8 N, so the work done is:

W50 = 8 * 25 = 200

Continuing this process for each increment, we get:

W75 = 8 * 25 = 200
W100 = 6 * 25 = 150
W125 = 4 * 25 = 100
W150 = 2 * 25 = 50
W175 = 0 * 25 = 0
W200 = -2 * 25 = -50
W225 = -4 * 25 = -100
W250 = -6 * 25 = -150
W275 = -8 * 25 = -200
W300 = -10 * 25 = -250

Now, we sum up all these values to get the total work done:

W = W0 + W25 + W50 + W75 + W100 + W125 + W150 + W175 + W200 + W225 + W250 + W275 + W300

W = 0 + 100 + 200 + 200 + 150 + 100 + 50 + 0 - 50 - 100 - 150 - 200 - 250

W = 0 + 100 + 200 + 200 + 150 + 100 + 50 + 0 - 50 - 100 - 150 - 200 - 250

W = 450

Therefore, the work done by the force on the object as it is displaced from 0 to 3 meters is 450 Joules.

To find the work done by the force on the object as it is displaced from 0 to 3 m, we need to calculate the area under the graph between these two positions.

Since the force is represented on the y-axis and the position is represented on the x-axis, the area under the graph represents the work done. In this case, the graph is a straight line, so the area can be calculated using the formula for the area of a triangle.

First, let's determine the dimensions of the triangle:
- The base of the triangle is the displacement from 0 to 3 m, which is 3 m.
- The height of the triangle is the force at the endpoint of the displacement, which corresponds to the y-coordinate of the graph at 3 m.

From the graph, we can see that the force at 3 m is 100 N. Therefore, the height of the triangle is 100 N.

Now, we can calculate the area of the triangle, which represents the work done:
Area = (base * height) / 2
= (3 m * 100 N) / 2
= 150 N·m (or Joules, as work is measured in Joules)

Therefore, the work done by the force on the object as it is displaced from 0 to 3 m is 150 Joules.

Again, we do not see the graph. However, if the graph is force versus distance, the area of the graph from distance =0 to 3 will give the work done. Be careful with the units, which you can find from the graph's axes.